Question

Find the point on the curve y² = 4x which is nearest to the point (2,-8) ​

Answer 1

Step-by-step explanation:

Let p(t², 2t ) be a point on y² = 4x nearest to the point (2,-8) then

AP² = (t²-2)² + (2t+8)²

Then , taking y = AP²

y = (t²-2)² + (2t+8)²

=(t⁴- 4t² +4)+(4t²+32t+64)

= t⁴ + 32t +68

dy/dx = 4t³ + 32 and d²y/dt² = 12t²

For max or minimum value of y

dy/dt = 0

=> 4(t³+8) = 0

=>(t+2)(t²-2t+4) = 0

=> t = &2

(t² -2t +4) = 0 has imaginary value of t )

for , t = -2 , d²y/dx² = 12(-2)2=48 >0

=> y is min , hence AP is min

P(t²,2t) = (4,-4) is the point on y²= 4x ,nearest given point A(2,-8)

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Answer 2

Answer:

refer to this attachment ✅✅